MHB ANSWER CHECK: Sum of Double Integrals involving Polar Conversion

[Pull and Twist]

Here is the given problem...

View attachment 6022

And I first approached it by drawing the xy footprint to get my theta and radius limits to convert to polar.

Then I overlooked the original xy function and pretty much took the area of that footprint (highlighted in green.) That gave me a very nice number.

Later I realized that I should have converted xy into polar as well as the object is probably intended to have varying density (highlighted in red.) With that approach I got the answer 15. Also a nice clean number.

Can anyone verify that the second answer is correct or am I approaching this the wrong way?

Thanks.

View attachment 6023

And here's the footprint I drew...
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[HallsofIvy]

Okay, why "ignore xy"?? Obviously that will change the answer!. The region over which you are integrating- what you call the "footprint"- is the region bounded below by the x-axis, y= 0, or \theta= 0 in polar coordinates, above by y= x, \theta= \frac{\pi}{4} in polar coordinates, on the left by the circle x^2+ y^2= 4, r= 2 in polar coordinates, and on the right by the circle x^2+ y^2= 16, r= 5 in polar coordinates. To integrate over that region in polar coordinates integrate \int_0^{\pi/4}\int_2^4 f(r, \theta) r dr d\theta where "f(r, \theta)" is xy in polar coordinates.

Yes, 15 is the correct answer!

Added for those who might be interested (and because I just can't resist):
In polar coordinates, x= r cos(\theta) and y= r sin(\theta) so xy= r^2 sin(theta)cos(\theta).

The integral becomes \int_0^{\pi/4}\int_2^4 r^2 sin(\theta)cos(\theta) (r dr d\theta). Since the integrand is just a product of functions of r, r^3, and \theta, sin(\theta)cos(\theta), and the limits of integration are constants, we can separate that integral as \left(\int_0^{\pi/4}sin(theta)cos(\theta)d\theta\right)\left(\int_2^4 r^3 dr\right).

The second integral is obviously \int_2^4 r^3 dr= \left[\frac{1}{4}r^4\right]_2^4= 64- 4= 60.

To do the first integral, let u= sin(\theta) so that du= cos(\theta)d\theta. When \theta= 0, u= sin(0)= 0 and when \theta= \pi/4, u= 1/\sqrt{2} so we have \int_0^{\pi/4} sin(theta)cos(\theta)d\theta= \int_0^{1/\sqrt{2}} u du= \left[\frac{1}{2}u^2\right]_0^{1/\sqrt{2}}= \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)= \frac{1}{4}.

So the original integral is \frac{60}{4}= 15

 

[Pull and Twist]

Okay, why "ignore xy"?? Obviously that will change the answer!. The region over which you are integrating- what you call the "footprint"- is the region bounded below by the x-axis, y= 0, or \theta= 0 in polar coordinates, above by y= x, \theta= \frac{\pi}{4} in polar coordinates, on the left by the circle x^2+ y^2= 4, r= 2 in polar coordinates, and on the right by the circle x^2+ y^2= 16, r= 5 in polar coordinates. To integrate over that region in polar coordinates integrate \int_0^{\pi/4}\int_2^4 f(r, \theta) r dr d\theta where "f(r, \theta)" is xy in polar coordinates.

Yes, 15 is the correct answer!

I didn't mean to ignore it... and obviously I realized I did after the fact... then I second guessed myself.

Thank you for verifying that I took the correct approach the second time around.